Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Problem 2.5: Find the electric field a distance $z$ above the center of a circular loop of radius $r$ which carries a uniform line charge $\lambda$.

This problem is in refereced here (with solution): http://teacher.pas.rochester.edu/PHY217/LectureNotes/Chapter2/LectureNotesChapter2.html

But it is solved using a trick. The trick is: okay we know there is symmetry so let us just take the magnitude and not use the vectors. The problem I have is I am unable to solve it without using the trick and I would appreciate someones help in doing the problem without the shortcut, using the vectors. When I do it the r compontent doesn't cancel.

Using polar coordinates here is the setup:

$$dE = \frac{1}{4 \pi \epsilon_0} \frac{\lambda}{\rho^2} ~ \hat{\rho} ~ r d \theta$$

$$\hat{\rho} = \frac{z \hat{k} + r \hat{r}}{\sqrt{z^2 + r^2}} $$

$$ E = \frac{\lambda}{4 \pi \epsilon_0} \int_0^{2 \pi} \frac{r z \hat{z} + r^2 \hat{r} }{(z^2 + r^2)^{3/2}} ~ d \theta$$

$$ E = \frac{\lambda}{4 \pi \epsilon_0} \int_0^{2 \pi} \frac{r z \hat{z} }{(z^2 + r^2)^{3/2}} ~ d \theta + \frac{\lambda}{4 \pi \epsilon_0} \int_0^{2 \pi} \frac{r^2 \hat{r} }{(z^2 + r^2)^{3/2}} ~ d \theta $$

If I now evaluate the $z$ component I get the correct answer. When I evaluate the $r$ component, shown below the $r$ doesn't not cancel the way it should. why not?

$$\frac{\lambda}{4 \pi \epsilon_0} \frac{r^2 2 \pi }{(z^2 + r^2)^{3/2}} - \frac{\lambda}{4 \pi \epsilon_0} \frac{r^2 0 }{(z^2 + r^2)^{3/2}} $$

$$ = \frac{\lambda}{4 \pi \epsilon_0} \frac{r^2 2 \pi }{(z^2 + r^2)^{3/2}} \hat{r}$$

share|improve this question

closed as too localized by dmckee May 31 '13 at 0:37

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

1  
note that $\hat{r}$ is not a constant vector! –  Ali May 28 '13 at 17:05
    
Duplicate : physics.stackexchange.com/questions/62637/… –  ABC May 29 '13 at 3:11

1 Answer 1

up vote 2 down vote accepted

Recall that $\hat r$ depends on $\theta$ in the following way: $$ \hat r = \cos\theta \,\hat x +\sin\theta \,\hat y $$ It follows that $$ \int_0^ {2\pi }\hat r \,d\theta = \hat x \int_0^ {2\pi}\cos\theta \, d\theta + \hat y \int_0^{2\pi}\sin\theta \,d\theta = 0+0 $$

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.